PLEASE ANSWER IT STEP BY STEP AND EXPLAIN REALLY WELL I AM A

**********************PLEASE ANSWER IT STEP BY STEP AND EXPLAIN REALLY WELL I AM A SLOW LEARNER******************

QUESTION 1

A circular disk of mass M=0.85 kg and radius R=0.22 m rotates in a horizontal plane with angular velocity w=6.0 rad/s about a vertical axis through its center .

A) Find the time required for the disk to make 4 complete revolutions.

B) A small clay ball of mass m=165g is now placed on the turntable ( while it is rotating ) at a distance a= 0.165 m from the rotation axis. Using angular momentum conservation, determine the new angular velocity of the system.

C) Calculate the total rotation kinetic energy of the system after the ball is placed on the disk . Is energy lost or gained when the ball is placed onthe disk?

QUESTION 2

A vertical post of mass M=16.0 kg and lenght L=1.80m is supported by a wire that has one end attached to the top of the post and the other end attached to the groun at a distance d=3.60m from the bottom of the post. A horizontal applied force Fapp=75.0N is exerted on the post at a point y=1.35m above the ground.

A) list 4 forces that act on the post. Make diagram of these forces that specifies the direction of each force and where it acts on the post.

B) Consider a rotation axis that passes through point P at the bottom of the post and is perpendicular to the plane of the page. By balancing the torques acting on the post with respect to this rotation axis, determine the tension in the wire.

C) By balancing the vertical and horizontal forces acting on the post, find the normal force (vertical force) and static frictional force( horizontal force) that the ground exerts on the post.

Solution

A) time = angle revolved / angular speed

t = (4 x 2pi) / (6) =4.19 sec

B) initial moment of inertia = M R^2 /2 = 0.85 x 0.22^2 /2 = 0.02057 kg m^2

final moment of inertia, If = 0.02057 + (0.165 x 0.165^2) =0.02506 kg m^2

using angular momentum (IW) conservation,

0.02057 x 6 = 0.02506 x wf

wf = 4.92 rad/s

C) KE \\= If wf^2 /2

     = (0.02506) (4.92^2) /2 = 0.303 J

energy is lost.

------------------------

2.

PLease post picture (Figure) for this.

Otherwise description is very confusing.